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I'm a newcomer to Quantum Computing and I'm currently working through a tutorial shown here.

At one point (second last section in Part II), there is an exercise that I'm really struggling with. I'm new to bra-ket notation and, while I can spot some identities given previously in the tutorial that might be useful for the exercise, I'm not able to make much headway.

The question is: Show that for any matrix $M$ and vector $|ψ⟩$, the following identity holds, expressing the length of $M|\psi⟩$:

$\|M|\psi\rangle\|^2 = \langle\psi|M^\dagger M |\psi\rangle.$

My rather poor attempt starts with recognising that $M^\dagger M = MM=M^2=I$, where $I$ is the identity matrix, substituting this into the original formula. This identity was given earlier in the tutorial. This gives me:

$\|M|\psi\rangle\|^2 = M^2\langle\psi|\psi\rangle.$ (1)

Then, noting another identity provided, that $\langle\psi|\psi\rangle = |||\psi\rangle||^2$ and substituting into (1) gives:

$\|M|\psi\rangle\|^2 = M^2|||\psi\rangle||^2$ (2)

At this point I'm basically stuck, I can't tell if I've provided proof, if I'm on the right track but not quite there, or if I'm totally wrong!

The only other thought I'd had is to use the identity:

$\langle\psi|M^\dagger = (M|\psi\rangle)^\dagger$,

but in honesty didn't really get any further than thinking I might need to use it.

I've been able to more or less follow the examples that the tutorial has worked through, but for this exercise I'm struggling even to understand how to proceed. Any advice would be much appreciated.

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  • $\begingroup$ The identity $M^\dagger M=M^2=I$ does not generally hold unless you assume that $M$ is a self-adjoint unitary. $\endgroup$
    – Condo
    Commented Apr 20, 2022 at 14:52
  • $\begingroup$ Why is $M^\dagger M = I$? The question said for any matrix $M$ not necessary unitary matrix $M$ $\endgroup$
    – KAJ226
    Commented Apr 20, 2022 at 14:52
  • $\begingroup$ Thankyou both for your replies. At least I know now that I'm on the wrong path. $\endgroup$
    – McKendrigo
    Commented Apr 20, 2022 at 14:58

1 Answer 1

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It follows from the definition of the norm on an inner-product space (that is a vector space equipped with an inner-product). For any vector $|\psi\rangle$ in an $n$-dimensional inner-product space $V$ we have by definition that $$\||\psi\rangle\|^2=\bigg\langle |\psi\rangle\bigg||\psi\rangle\bigg\rangle=\langle \psi|\psi\rangle.$$

If $M$ is any matrix $n\times n$ (i.e. a linear operator acting on $V$, written $M: V\rightarrow V$) then we have $$M|\psi\rangle=|\phi\rangle$$ for some $|\phi\rangle\in V$. The norm of $M|\psi\rangle=|\phi\rangle$ is then $$\|M|\psi\rangle\|^2=\||\phi\rangle\|^2=\langle \phi|\phi\rangle=\langle M\psi|M\psi\rangle.$$

Finally, for any linear operator $M$ acting on an inner-product space we define the adjoint as the operator $M^\dagger$ which satisfies the equation $$\langle Mv|w\rangle=\langle v|M^\dagger |w\rangle,$$ for all $|v\rangle,|w\rangle\in V$.

Thus, using our definition of the adjoint we see that $$\|M|\psi\rangle\|^2=\langle M\psi|M\psi\rangle=\langle\psi|M^\dagger M|\psi\rangle,$$ as desired.

I would also recommend reviewing the Bra-ket Wikepdia page, especially the section on common pitfalls with the notation.

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  • $\begingroup$ Thank you very much for your detailed answer. I must admit, a lot of that is going over my head right now but I will try to understand it. $\endgroup$
    – McKendrigo
    Commented Apr 20, 2022 at 15:26
  • $\begingroup$ The part of the answer I'm trying to get to grips with is the adjoint. ...we define the adjoint as the operator M† which satisfies the equation $⟨Mv|w⟩=⟨v|M^†w⟩,$ (1) for all $v,w∈V.$ Thus, using our definition of the adjoint we see that $∥M|ψ⟩∥^2=⟨Mψ|Mψ⟩=⟨ψ|M^†M|ψ⟩$ (2) Now, I'm trying to get from point (1) to point (2). This is my best attempt: Let v=$\psi$ and w=$M\psi$ and substituting into (1). Thus:$⟨Mv|w⟩=⟨M\psi|M\psi⟩=⟨v|M^†w⟩=⟨\psi|M^†M\psi⟩$ Almost looks right, but I can't understand why I have $M\psi⟩$ and not $M|\psi⟩$ at the very end. $\endgroup$
    – McKendrigo
    Commented Apr 20, 2022 at 16:23
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    $\begingroup$ It is just a clash of the bra-ket notation with conventional mathematical notation, one usually accepts that $|M\psi \rangle=M|\psi\rangle$. $\endgroup$
    – Condo
    Commented Apr 20, 2022 at 18:16
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    $\begingroup$ Using Bra-ket notation you could equivalently use the definition of the adjoint as $\langle M\psi|=\langle \psi|M^\dagger$. Perhaps that's simpler. $\endgroup$
    – Condo
    Commented Apr 20, 2022 at 18:25
  • $\begingroup$ Thank you @Condo, that has been very helpful $\endgroup$
    – McKendrigo
    Commented Apr 21, 2022 at 12:34

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